Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 10 - Rotational Kinematics and Energy - Problems and Conceptual Exercises: 7

Answer

$\omega=1.99\times 10^{-7} rad/s=1.9\times10^{-6}~rev/min$

Work Step by Step

To find the angular speed, use the definition $$\omega=\frac{\Delta \theta}{\Delta t}$$ The time for one full revolution is 365.25 days. Convert this to seconds using dimensional analysis. $$365.25 days \times \frac{24hr}{1day} \times \frac{60min}{1hr} \times \frac{60s}{1min}$$ $$=3.15\times 10^{7} s$$ Substituting known values of $\Delta \theta=2\pi rad$ (one full revolution) and $\Delta t=3.15\times 10^7m/s$ yields an angular speed of $$\omega=\frac{2\pi rad}{3.15\times 10^7s}=1.99\times 10^{-7} rad/s$$ We convert to $rev/min$: $1.99\times10^{-7}~rad/s*\frac{1~rev}{2\pi~rad}*\frac{60~s}{1~min}=1.9\times10^{-6}~rev/min$
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